Exercise · ZW-CONVERGENCE-WITH-LIPSCHITZ-GRADIENT-007 · ★★☆ Intermediate

For \( f(x)=x^2 \) on \( \mathbb{R} \), what is the Lipschitz constant \( L \) of the gradient?

Choose one option.

Show answer
  • C. \( L=2 \)

Explanation

Since \( \nabla f(x)=2x \), we have \( |\nabla f(x)-\nabla f(y)|=2|x-y| \), so \( L=2 \). Choosing \( L=1 \) forgets the factor 2 from the derivative.

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Review the concept: Convergence with Lipschitz gradient

Result guaranteeing that gradient descent with a suitable step size converges for differentiable functions whose gradient \( \nabla f \) is Lipschitz continuous with constant \( L \).

\[f(x_k)-f^* \le \frac{L\|x_0-x^*\|^2}{2k}\]

More practice problems: convergence with lipschitz gradient

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